$Let (x_{0}, y_{0}) be solution of the following equations (2 x)^{\ln 2} = (3 y)^{\ln 3}, 3^{\ln x} = 2^{\ln y}$ $Then x_{0} is$

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$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$(2 x)^{\ln 2} = (3 y)^{\ln 3} - - - - - (1)$

$3^{\ln x} = 2^{\ln y} - - - - (2)$

In (2), taking log both sides,

$\begin{array}{l} (\log x) (\log 3) = (\log y) \log 2 \\ \Rightarrow \log y = \frac{(\log x) (\log 3)}{\log 2} - - - - - (3) \end{array}$

In (1) taking log both sides

$\begin{array}{l} (\log 2) {\log 2 + \log x} = \log 3 {\log 3 + \log y} \\ (\log 2)^{2} + (\log 2) (\log x) = (\log 3)^{2} + \frac{(\log 3)^{2} (\log x)}{\log 2} from (3) \end{array}$

$\begin{array}{l} \Rightarrow (\log 2)^{2} - (\log 3)^{2} = \frac{(\log 3)^{2} - (\log 2)^{2}}{\log 2} (\log x) \\ \Rightarrow - \log 2 = \log x \\ \Rightarrow x = \frac{1}{2} \Rightarrow x_{0} = \frac{1}{2} \end{array}$